## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$\dfrac{ 2 \pi }{3}$
We are given: $\cos^{-1} [\sin (\dfrac{7 \pi}{ 6})]$ We know that the trigonometric function $\cos^{-1}$ has domain $[-1,1]$ and range: $[0, \pi]$. Here, $\cos^{-1} (\dfrac{-1}{2})=\dfrac{7\pi}{6}$, because $\cos{(\dfrac{7\pi}{6})}=\dfrac{-1}{2}$ and $\dfrac{7\pi}{6}$ lies in the range of $\cos^{-1}{x}$ . Therefore, we have: $\cos^{-1} [\sin (\dfrac{7 \pi}{ 6})]=\cos^{-1} ({\dfrac{-1}{2}})=\dfrac{ 2 \pi }{3}$